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I have a problem in finding the reaction force of the pinned-fixed structure. If I take moment at both ends, the reaction forces at A and B should be the same and the answer is wrong. How to derive the reaction forces and the bending moment? enter image description here

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3 Answers 3

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Since the moment at the fixed end $A$ is given, then you can sum the moment about $A$ or $B$ to solve the reactions. Let's start with $\sum M_A = 0$ first.

$R_B*L + \dfrac{3PL}{16} - \dfrac{PL}{2} = 0$

$R_B = \dfrac{P}{2} - \dfrac{3P}{16} = \dfrac{5P}{16}$

Now, $\sum M_B = 0$,

$R_A*L - \dfrac{3PL}{16} - \dfrac{PL}{2} = 0$

$R_A = \dfrac{3P}{16} + \dfrac{P}{2} = \dfrac{11P}{16}$

Finally, $\sum F_y = 0$

$R_A + R_B - P = \dfrac{11P}{16} + \dfrac{5P}{16} - P = P - P = 0$, check!

Note, when taking the moment about a joint, you can take either CW or CCW as positive rotation but must be consistent throughout that operation, then the result will be carrying the correct sign.

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  • $\begingroup$ Thanks for your answer! However, how can I derive the given moment if I do not know the formula? $\endgroup$
    – kk6
    Apr 13 at 4:05
  • $\begingroup$ how to come up with 3PL/16? $\endgroup$
    – kk6
    Apr 13 at 4:10
  • $\begingroup$ The simplest way, look up your structural analysis textbook, there must be a page or two providing the FEM for the single span beam. The second method is to release the support at "B", calc the deflection at the freed end "B", and place a unit load to close the gap (I assume you know the consistent displacement method), then figure out R_B by proportion, From here, with R_B is known, this becomes a simple cantilever problem. $\endgroup$
    – r13
    Apr 13 at 4:38
  • $\begingroup$ May I ask a stupid question? Why isn't taking moment at A possible in this question? P(L/2)+RB(L)=0 As i got RB= P/2 from this method. $\endgroup$
    – kk6
    Apr 13 at 4:41
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    $\begingroup$ For this system, there are 4 unknown forces (3 at support "A", and 1 at "B"), but there are only 3 equilibrium equations available, so, this is a structurally indeterminate beam to the first degree (4 unknowns - 3 equations), which can't be solved using the method for simply supported beam. For this case, you need to find one more equation or condition, so the number of equations = the number of unknowns. (Note, for simply supported beam, there are 3 unknowns with 3 available equilibrium equations) $\endgroup$
    – r13
    Apr 13 at 4:56
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A shorthand calculation of the left-hand moment is:

We first consider a fixed-fixed beam loaded at the center with P.

The fixed end moments are

$$M=-PL/8$$

Now we release the moment at righthand support returning it to pin support with zero moment,$M_B=0.$

We know half of this moment, $1/2(-PL/8)= -PL/16 \ $ will be transferred to the lefthand support, $A.$

$$\therefore \ -PL/8+(-PL/16)=-3PL/16$$

That is the moment you have in your question. I used sign convention as a positive moment creating tension on the bottom fibers, but it doesn't matter!

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  • $\begingroup$ I like this. It feels like a very simplified version of the Hardy Cross (moment distribution) method. $\endgroup$
    – hazzey
    Apr 13 at 10:54
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A slightly more convoluted (but completely general) procedure is to work out the differential equations given by Euler-Bernoulli beam theory.

The first step is to replace all unknown reactions (the fixed support at A in this case) by an equivalent unknown load that will be treated sepparately. This can be done because the beam in question is considered linear elastic.

In this case your system becomes as follows System separation

Now you proceed to determine the support reaction and bending moments for both these systems. Since both are simple isostatic cases I'll just state the results.

$$ M_1(x)= \begin{cases} \frac{P}{2}x & \text{if }x\leq\frac{L}{2}\\ \frac{P}{2}x - P(x-\frac{L}{2}) & \text{if } x> \frac{L}{2} \end{cases} \qquad M_2(x)=\frac{M}{L}x-M $$

Now, for a beam with constant elastic properties $E$ and $I$ you can write the curvature as $$ \varphi(x)=\frac{M(x)}{EI} $$ Ane Euler-Bernoulli beam theory tells us that $$ \theta(x)=\int\varphi(x)\,dx \quad \text{and}\quad v(x)=\int\theta(x)\,dx $$ This is technically the solution to a $2^{nd}$ degree ODE, but it's so simple it can be solved by direct integration. Results for deflection $v(x)$ are stated below $$ v_1(x)= \frac{1}{EI} \begin{cases} \frac{Px^3}{12}+C_ax+D_a & \text{if }x\leq\frac{L}{2} \\ \frac{PLx^2}{4} - \frac{Px^3}{12}+C_bx+D_b & \text{if } x> \frac{L}{2} \end{cases} \qquad v_2(x)=\frac{1}{EI}\left(\frac{M}{L}\frac{x^3}{6}-M\frac{x^2}{2} +C_2x+D_2\right) $$ To find the integration constants we have the following conditions $$ v_1(0)=0\qquad v_1(\frac{L}{2}^-)=v_1(\frac{L}{2}^+) \qquad \theta_1(\frac{L}{2}^-)=\theta_1(\frac{L}{2}^+) \qquad v_1(L)=0 \\ v_2(0)=0 \qquad v_2(L)=0 $$ Solving for these constants nets you $$ v_1(x)=\frac{1}{EI} \begin{cases} \frac{Px^3}{12}-\frac{PL^2}{16}x & \text{if }x\leq\frac{L}{2} \\ \frac{PLx^2}{4} - \frac{Px^3}{12}-\frac{3PL^2}{16}x+\frac{PL^3}{48} & \text{if } x> \frac{L}{2} \end{cases} \qquad v_2(x)=\frac{1}{EI}\left(\frac{M}{L}\frac{x^3}{6}-M\frac{x^2}{2} +\frac{ML}{3}x\right) $$ With this, you need only apply the condition defining the fixed support at A. this condition is $\theta_1(0)+\theta_2(0)=0$, using the value of $M$ to enforce this equality. Doing the algebra leads to

$$ M=\frac{3PL}{16} $$ Exactly as the value you got on the original prompt.

This looks like a lot of algebra, but the main advantage is that the procedure is completely general. Following the same steps you can solve a beam having a any end support conditions, supports at the mid-span, concentrated or distributed loads, etc. Furthermore, this allows to calculate bending moment and shear at any point, which is also an important consideration. With some work you can write a program that solves this numerically, needing to work out all of this only once.

As a final note, if you plot the deflections $v_i(x)$, positive values point downward.

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